Chapter 8: Q. 59 (page 702)

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx$

Short Answer

Expert verified

$\int_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} 3^{k}}{k!} (\frac{1 - {(0.5)}^{2 k + 3}}{2 k + 3})$

Step by step solution

Step 1. Given information is:

$\int_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx$

Step 2. Definite integral

$From Q 49 . Maclaurin series for f (x) = x^{2} e^{- 3 x^{2}} is x^{2} e^{- 3 x^{2}} = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} 3^{k}}{k!} {(x)}^{2 k + 2} Also, F = \int f F (x) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} 3^{k}}{k!} (\frac{{(x)}^{2 k + 3}}{2 k + 3}) Adding the limits, F (x) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} 3^{k}}{k!} {[\frac{{(x)}^{2 k + 3}}{2 k + 3}]}_{0.5}^{1} = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} 3^{k}}{k!} (\frac{{(1)}^{2 k + 3} - {(0.5)}^{2 k + 3}}{2 k + 3}) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} 3^{k}}{k!} (\frac{1 - {(0.5)}^{2 k + 3}}{2 k + 3})$

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Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx$

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Definite integral

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Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.∫0.51x2e-3x2dx

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Definite integral

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Calculus

Mechanics Maths

Logic and Functions

Probability and Statistics

Statistics

Study anywhere. Anytime. Across all devices.

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx$